In a large organisation, only 1 out of every 1,500 emails contains a harmful attachment. An…

Let $H$ be the event that an email is harmful, and $S = \bar{H}$ be the event that it is safe. Given $P(H) = \frac{1}{1500}$. The probability that the email is safe is $P(S) = 1 - P(H) = 1 - \frac{1}{1500} = \frac{1499}{1500}$.

Let $D$ denote the event that the filter marks an email as dangerous, and $\bar{D}$ denote the event that the filter marks it as safe. Given $P(D|H) = 0.97$ (filter correctly marks harmful email dangerous). The probability that given the email is harmful, the filter marks it safe is: $P(\bar{D}|H) = 1 - P(D|H) = 1 - 0.97 = 0.03$.

We are given that an email has been flagged as dangerous ($D$). We want to find $P(S|D)$: $P(D|H) = 0.97$, $P(H) = \frac{1}{1500}$. $P(D|S) = 1 - P(\text{safe}|S) = 1 - 0.97 = 0.03$, $P(S) = \frac{1499}{1500}$. Total probability of flagging as dangerous: $P(D) = P(D|H)P(H) + P(D|S)P(S) = 0.97\left(\frac{1}{1500}\right) + 0.03\left(\frac{1499}{1500}\right) = \frac{0.97 + 44.97}{1500} = \frac{45.94}{1500} \approx 0.030627$. By Bayes' theorem: $P(S|D) = \frac{P(D|S)P(S)}{P(D)} = \frac{0.03 \times \frac{1499}{1500}}{\frac{45.94}{1500}} = \frac{44.97}{45.94} \approx 0.9789 \approx 0.98$.